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L. R. Flores CastilloCUHK January 28, 2015 Symmetry, conservation laws, groups 1917: Emmy Noether’s theorem: Every symmetry yields a conservation law Conversely, every conservation law reflects an underlying symmetry 6 A “symmetry” is an operation on a system that leaves it invariant. i.e., it transforms it into a configuration indistinguishable from the original one. The set of all symmetry operations on a given system forms a group: Closure: If a and b in the set, so is ab Identity: there is an element I s.t. aI = Ia = a for all elements a. Inverse: For every element a there is an inverse, a -1, such that aa -1 = a -1 a = I Associativity: a(bc) = (ab)c if commutative, the group is called Abelian

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L. R. Flores CastilloCUHK January 28, 2015 Addition of Angular Momenta Example 2. Two spin-½ states combine to give spin 1 and spin 0. Find the explicit Clebsch-Gordan decomposition. 22 Equivalently (solving for states with j=0,1): N.B.: we could have read these coefficients from the Clebsch-Gordan table (it works both ways)

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L. R. Flores CastilloCUHK January 28, 2015 Spin ½ Most important case (p, n, e, all quarks, all leptons) Illustrative for other cases For s=½, 2 states: m s =½ (“spin up”,  ) or m s = –½ (“spin down”,  ) Better notation: Spinors –two-component column vectors: “a particle of spin ½ can only exist in one of these states” Wrong! its general state is Where α and β are complex numbers, and 23

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L. R. Flores CastilloCUHK January 28, 2015 Spin ½ The measurement of S z can only return +½ħ and -½ħ with probabilities |α| 2 and |β| 2, respectively If we now measure S x or S y on a particle in this state, –what possible results may we get? +½ħ and -½ħ –what is the probability of each result? 24

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L. R. Flores CastilloCUHK January 28, 2015 Spin ½ In general, in QM: Construct the matrix Â representing the observable A The allowed values of A are the eigenvalues of Â –Eigenvalues: e i, eigenvectors: Write the state of the system as a linear combination of the eigenvectors of Â The probability that a measurement of A would yield the value e i is |c i | 2 25

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L. R. Flores CastilloCUHK January 28, 2015 Spin ½ Eigenvalues of S x are ±ħ/2, corresponding to normalized eigenvectors: any spinor can be written as a linear combination of these eigenvectors: by choosing 26 Construct matrix Â representing observable A Allowed values of A are the eigenvalues of Â Write state as linear combination of these eigenvectors The probability to measure e i is |c i | 2

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L. R. Flores CastilloCUHK January 28, 2015 Spin ½ In terms of the Pauli spin matrices: the spin operators can be written as Effect of rotations on spinors: where θ is a vector pointing along the axis of rotation, and its magnitude is the angle of rotation. 27

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L. R. Flores CastilloCUHK January 28, 2015 Spin ½ These matrices U(θ) are Unitary, of determinant 1. i.e., they constitute the group SU(2) Spin-½ particles transform under rotations according to the two-dimensional representation of the group SU(2) Particles of spin 1, described by vectors, transform under the three-dimensional representation of SU(2) Particles of spin 3/2: described by 4-component objects, transform under the 4d representation of SU(2) Why SU(2)? the group is very similar (homomorphic) to the SO(3) (the group of rotations in three dimensions). Particles of different spin belong to different representations of the rotation group. 28

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L. R. Flores CastilloCUHK January 28, 2015 Flavor Symmetries Shortly after the discovery of the neutron (1932) Heisenberg observed that the neutron is very similar to the proton –m p = MeV/c 2 ; m n = MeV/c 2. Two “states” of the same particle? (the nucleon) Maybe the mass difference was related to the charge? (it would be the other way around: p would be heavier) The nuclear forces on them are likely identical Implementation: 29

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L. R. Flores CastilloCUHK January 28, 2015 Flavor symmetries Horizontal rows: very similar masses but different charges We assign an isospin I to each multiplet And a particular I 3 to each member of the multiplet: Pion: I=1 : 32

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L. R. Flores CastilloCUHK January 28, 2015 Flavor symmetries It has dynamical implications; for example: Two nucleons (hence I=1) can combine into –a symmetric isotriplet: – an antisymmetric isosinglet –Experimentally, p & n form a single bound state (the deuteron) –There is no bound state of two protons or two neutrons –Therefore, the deuteron must be the isosinglet (otherwise all three states would need to occur). –There should be a strong attraction in the I=0 channel, and not in the I=1 channel. 36

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L. R. Flores CastilloCUHK January 28, 2015 Flavor symmetries Nucleon-nucleon scattering: From the one in the middle, only the I=1 combination contributes. As a result, the scattering amplitudes are in the ratio: and the cross sections (~ square of s.a.’s): 37

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L. R. Flores CastilloCUHK January 28,

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L. R. Flores CastilloCUHK January 28, 2015 Flavor symmetries In the 50’s, as more particles were found, it was tempting to extend this idea, but it became increasingly hard to argue that they were different states of the same particle The Λ, Ξ’s and Σ’s could be regarded as a supermultiplet, as if they belonged in the same representation of an enlarged symmetry group SU(2) of isospin would then be a subgroup, but what was the larger group? The Eightfold way was Gell-Mann’s solution: The symmetry group is SU(3), the octets are 8D representations of SU(3) the decuplet a 10D representation, etc. 39

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Backup 40

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L. R. Flores CastilloCUHK January 28, 2015 Eigenvectors and eigenvalues A nonzero vector χ is called an eigenvector of a matrix M if M χ = λ χ for some number λ (the eigenvalue). (notice that any multiple of χ is also an eigenvector of M, with the same eigenvalue) 41